What is the domain of y= 1/(4x^2-4x-3) ?

The domain of y= 1/(4x^2-4x-3) is x 3/2

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domain of y= 1/(4x^2-4x-3)

domain

y = \frac{1}{4 x ^{2} - 4 x - 3}

x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{3}{2} or x > \frac{3}{2}

Interval Notation

(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \frac{3}{2}) \cup (\frac{3}{2}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{3}{2}, x = - \frac{1}{2}

The function domain

x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{3}{2} or x > \frac{3}{2}

\frac{x ^{2} - 4 x - 12}{x + 2}

f (x) = 4 x^{2} + 1

f (x) = 16 - x^{2} + x - 3

f (x) = lo g_{10} (3) (x) + 2

\frac{x ^{2} - 7 x + 10}{x ^{2} - 5 x + 6}

What is the domain of y= 1/(4x^2-4x-3) ?
The domain of y= 1/(4x^2-4x-3) is x<-1/2 \lor-1/2 <x< 3/2 \lor x> 3/2